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Question

The sum of coefficients of all the integral powers of  $$x$$ in the expansion of $$(1+2\sqrt{x})^{40}$$ is


A
340+1
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B
3401
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C
12 (3401)
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D
12 (340+1)
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Solution

The correct option is D $$\displaystyle\frac{1}{2}$$ $$(3^{40}+1)$$
$$(1+2\sqrt{x})^{40}=1+\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}+\:^{40}C_{3}(2\sqrt{x})^{3}+...+\:^{40}C_{40}(2\sqrt{x})^{40}$$
Hence, all the odd terms will have integral powers of $$x$$.
Substituting $$x=1$$, we get
$$3^{40}=a_{0}+a_{1}+a_{2}+a_{3}...+a_{40}$$ ...(i)
Now consider,
$$(1-2\sqrt{x})^{40}=1-\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}-\:^{40}C_{3}(2\sqrt{x})^{3}...+\:^{40}C_{40}(2\sqrt{x})^{40}$$
Taking $$x$$ as $$1$$, we get
$$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{40}$$ ...(ii)
Adding i and, ii we get
$$3^{40}+1=2[a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}]$$
$$\dfrac{3^{40}+1}{2}=a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}$$
Hence the sum of the coefficients, of integral powers of $$x$$ will be
$$\dfrac{3^{40}+1}{2}$$
Hence, option D is correct

Mathematics

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